if x=a(y+z), y= b(z+x) & z= c(x+y) then, prove that, ab+bc+ca+2abc=1.   no

Expert Answers

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`a = x/(y+z)`

`b = y/(z+x)`

`c = z/(x+y)`

 

`ab = (xy)/((y+z)(z+x))`

`bc = (yz)/((z+x)(x+y))`

`ca = (xz)/((y+z)(x+y))`

`abc = (xyz)/((y+z)(x+y)(x+z))`

 

`ab+bc+ca+2abc`

`=(xy)/((y+z)(z+x))+(yz)/((z+x)(x+y))+(xz)/((y+z)(x+y))+(2xyz)/((y+z)(x+y)(x+z))`

Now we get  `(x+y)(y+z)(z+x)` as the common denominator.

`= ((xy(x+y)+yz(y+z)+xz(x+z)+2xyz))/((y+z)(x+y)(x+z))`

`= (x^2y+xy^2+y^2z+z^2y+x^2z+z^2x+2xyz)/((y+z)(x+y)(x+z))`

`= (x^2y+2xyz+z^2y+xy^2+zy^2+x^2z+z^2x)/((y+z)(x+y)(x+z))`

`= (y(x^2+2xz+z^2)+y^2(x+z)+zx(x+z))/((y+z)(x+y)(x+z))`

(x+z) term is common for all.

`= ((x+z)(y(x+z)+y^2+zx)/((y+z)(x+y)(x+z))`

`= ((x+z)(yx+yz+y^2+zx))/((y+z)(x+y)(x+z))`

`= ((x+z)(y(x+y)+z(x+y)))/((y+z)(x+y)(x+z))`

`= ((x+z)(x+y)(y+z))/((y+z)(x+y)(x+z))`

`= 1`

 

So the answer is proved.

 

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